如何在python中找到兩個日期時間對象之間的時差？

Question

如何判斷兩個datetime對象之間的時差（以分鍾為單位）？

Answer 1

>>> import datetime
>>> first_time = datetime.datetime.now()
>>> later_time = datetime.datetime.now()
>>> difference = later_time - first_time
datetime.timedelta(0, 8, 562000)
>>> seconds_in_day = 24 * 60 * 60
>>> divmod(difference.days * seconds_in_day + difference.seconds, 60)
(0, 8)      # 0 minutes, 8 seconds

從第一個時間difference = later_time - first_time減去以后的時間difference = later_time - first_time創建一個僅保留差異的日期時間對象。 在上面的示例中，它是 0 分 8 秒和 562000 微秒。

Answer 2

使用日期時間示例

>>> from datetime import datetime
>>> then = datetime(2012, 3, 5, 23, 8, 15)        # Random date in the past
>>> now  = datetime.now()                         # Now
>>> duration = now - then                         # For build-in functions
>>> duration_in_s = duration.total_seconds()      # Total number of seconds between dates

年限

>>> years = divmod(duration_in_s, 31536000)[0]    # Seconds in a year=365*24*60*60 = 31536000.

持續時間（天）

>>> days  = duration.days                         # Build-in datetime function
>>> days  = divmod(duration_in_s, 86400)[0]       # Seconds in a day = 86400

持續時間（小時）

>>> hours = divmod(duration_in_s, 3600)[0]        # Seconds in an hour = 3600

持續時間（分鍾）

>>> minutes = divmod(duration_in_s, 60)[0]        # Seconds in a minute = 60

持續時間（以秒為單位）

[!] 請參閱本文底部有關以秒為單位使用持續時間的警告

>>> seconds = duration.seconds                    # Build-in datetime function
>>> seconds = duration_in_s

以微秒為單位的持續時間

[!] 請參閱本文底部有關使用以微秒為單位的持續時間的警告

>>> microseconds = duration.microseconds          # Build-in datetime function

兩個日期之間的總持續時間

>>> days    = divmod(duration_in_s, 86400)        # Get days (without [0]!)
>>> hours   = divmod(days[1], 3600)               # Use remainder of days to calc hours
>>> minutes = divmod(hours[1], 60)                # Use remainder of hours to calc minutes
>>> seconds = divmod(minutes[1], 1)               # Use remainder of minutes to calc seconds
>>> print("Time between dates: %d days, %d hours, %d minutes and %d seconds" % (days[0], hours[0], minutes[0], seconds[0]))

或者干脆：

>>> print(now - then)

---

編輯 2019由於此答案已引起關注，我將添加一個功能，該功能可能會簡化某些人的使用

from datetime import datetime

def getDuration(then, now = datetime.now(), interval = "default"):

    # Returns a duration as specified by variable interval
    # Functions, except totalDuration, returns [quotient, remainder]

    duration = now - then # For build-in functions
    duration_in_s = duration.total_seconds() 
    
    def years():
      return divmod(duration_in_s, 31536000) # Seconds in a year=31536000.

    def days(seconds = None):
      return divmod(seconds if seconds != None else duration_in_s, 86400) # Seconds in a day = 86400

    def hours(seconds = None):
      return divmod(seconds if seconds != None else duration_in_s, 3600) # Seconds in an hour = 3600

    def minutes(seconds = None):
      return divmod(seconds if seconds != None else duration_in_s, 60) # Seconds in a minute = 60

    def seconds(seconds = None):
      if seconds != None:
        return divmod(seconds, 1)   
      return duration_in_s

    def totalDuration():
        y = years()
        d = days(y[1]) # Use remainder to calculate next variable
        h = hours(d[1])
        m = minutes(h[1])
        s = seconds(m[1])

        return "Time between dates: {} years, {} days, {} hours, {} minutes and {} seconds".format(int(y[0]), int(d[0]), int(h[0]), int(m[0]), int(s[0]))

    return {
        'years': int(years()[0]),
        'days': int(days()[0]),
        'hours': int(hours()[0]),
        'minutes': int(minutes()[0]),
        'seconds': int(seconds()),
        'default': totalDuration()
    }[interval]

# Example usage
then = datetime(2012, 3, 5, 23, 8, 15)
now = datetime.now()

print(getDuration(then)) # E.g. Time between dates: 7 years, 208 days, 21 hours, 19 minutes and 15 seconds
print(getDuration(then, now, 'years'))      # Prints duration in years
print(getDuration(then, now, 'days'))       #                    days
print(getDuration(then, now, 'hours'))      #                    hours
print(getDuration(then, now, 'minutes'))    #                    minutes
print(getDuration(then, now, 'seconds'))    #                    seconds

警告：關於內置 .seconds 和 .microseconds 的警告
datetime.seconds和datetime.microseconds的上限分別為 [0,86400) 和 [0,10^6)。

如果 timedelta 大於最大返回值，則應謹慎使用它們。

例子：

end是start后 1 小時和 200 微秒：

>>> start = datetime(2020,12,31,22,0,0,500)
>>> end = datetime(2020,12,31,23,0,0,700)
>>> delta = end - start
>>> delta.microseconds
RESULT: 200
EXPECTED: 3600000200

end是start后的 1d 和 1h：

>>> start = datetime(2020,12,30,22,0,0)
>>> end = datetime(2020,12,31,23,0,0)
>>> delta = end - start
>>> delta.seconds
RESULT: 3600
EXPECTED: 90000

Answer 3

Python 2.7 的新功能是timedelta實例方法.total_seconds() 。 從 Python 文檔中，這相當於(td.microseconds + (td.seconds + td.days * 24 * 3600) * 10**6) / 10**6 。

參考： http : //docs.python.org/2/library/datetime.html#datetime.timedelta.total_seconds

>>> import datetime
>>> time1 = datetime.datetime.now()
>>> time2 = datetime.datetime.now() # waited a few minutes before pressing enter
>>> elapsedTime = time2 - time1
>>> elapsedTime
datetime.timedelta(0, 125, 749430)
>>> divmod(elapsedTime.total_seconds(), 60)
(2.0, 5.749430000000004) # divmod returns quotient and remainder
# 2 minutes, 5.74943 seconds

Answer 4

只需從另一個中減去一個。 您將獲得一個具有差異的timedelta對象。

>>> import datetime
>>> d1 = datetime.datetime.now()
>>> d2 = datetime.datetime.now() # after a 5-second or so pause
>>> d2 - d1
datetime.timedelta(0, 5, 203000)

您可以將dd.days 、 dd.seconds和dd.microseconds轉換為分鍾。

Answer 5

如果a ， b是日期時間對象，那么在 Python 3 中找到它們之間的時差：

from datetime import timedelta

time_difference = a - b
time_difference_in_minutes = time_difference / timedelta(minutes=1)

在早期的 Python 版本上：

time_difference_in_minutes = time_difference.total_seconds() / 60

如果a , b是原始日期時間對象，例如由datetime.now()返回，那么如果對象表示具有不同 UTC 偏移量的本地時間，例如，圍繞 DST 轉換或過去/未來日期，則結果可能是錯誤的。 更多詳細信息：查找日期時間之間是否過去了 24 小時 - Python 。

要獲得可靠的結果，請使用 UTC 時間或時區感知日期時間對象。

Answer 6

使用divmod：

now = int(time.time()) # epoch seconds
then = now - 90000 # some time in the past

d = divmod(now-then,86400)  # days
h = divmod(d[1],3600)  # hours
m = divmod(h[1],60)  # minutes
s = m[1]  # seconds

print '%d days, %d hours, %d minutes, %d seconds' % (d[0],h[0],m[0],s)

Answer 7

這就是我如何獲得兩個 datetime.datetime 對象之間經過的小時數：

before = datetime.datetime.now()
after  = datetime.datetime.now()
hours  = math.floor(((after - before).seconds) / 3600)

Answer 8

只是認為提及有關 timedelta 的格式可能會很有用。 strptime() 根據格式解析表示時間的字符串。

from datetime import datetime

datetimeFormat = '%Y/%m/%d %H:%M:%S.%f'    
time1 = '2016/03/16 10:01:28.585'
time2 = '2016/03/16 09:56:28.067'  
time_dif = datetime.strptime(time1, datetimeFormat) - datetime.strptime(time2,datetimeFormat)
print(time_dif)

這將輸出：0:05:00.518000

Answer 9

只查找天數：timedelta 有一個“天”屬性。 您可以簡單地查詢。

>>>from datetime import datetime, timedelta
>>>d1 = datetime(2015, 9, 12, 13, 9, 45)
>>>d2 = datetime(2015, 8, 29, 21, 10, 12)
>>>d3 = d1- d2
>>>print d3
13 days, 15:59:33
>>>print d3.days
13

Answer 10

我使用這樣的東西：

from datetime import datetime

def check_time_difference(t1: datetime, t2: datetime):
    t1_date = datetime(
        t1.year,
        t1.month,
        t1.day,
        t1.hour,
        t1.minute,
        t1.second)

    t2_date = datetime(
        t2.year,
        t2.month,
        t2.day,
        t2.hour,
        t2.minute,
        t2.second)

    t_elapsed = t1_date - t2_date

    return t_elapsed

# usage 
f = "%Y-%m-%d %H:%M:%S+01:00"
t1 = datetime.strptime("2018-03-07 22:56:57+01:00", f)
t2 = datetime.strptime("2018-03-07 22:48:05+01:00", f)
elapsed_time = check_time_difference(t1, t2)

print(elapsed_time)
#return : 0:08:52

Answer 11

這將給出以秒為單位的差異（然后除以 60 得到分鍾）：

import time
import datetime

t_start = datetime.datetime.now()

time.sleep(10)

t_end = datetime.datetime.now()
elapsedTime = (t_end - t_start )

print(elapsedTime.total_seconds())

輸出：

10.009222

這是我認為最簡單的方法，您無需擔心精度或溢出。

例如，使用elapsedTime.seconds會損失很多精度（它返回一個整數）。 此外，正如這個答案指出的那樣， elapsedTime.microseconds的上限為 10^6。 因此，例如，對於 10 秒的sleep() ， elapsedTime.microseconds給出8325 （這是錯誤的，應該在10,000,000左右）。

Answer 12

這是為了找出當前時間和上午 9.30 之間的差異

t=datetime.now()-datetime.now().replace(hour=9,minute=30)

Answer 13

要獲得hour 、 minute和second ，您可以這樣做

>>> import datetime
>>> first_time = datetime.datetime.now()
>>> later_time = datetime.datetime.now()
>>> difference = later_time - first_time
>>> m,s = divmod(difference.total_seconds(), 60)
>>> print("H:M:S is {}:{}:{}".format(m//60,m%60,s)

Answer 14

這是我使用mktime 的方法。

from datetime import datetime, timedelta
from time import mktime

yesterday = datetime.now() - timedelta(days=1)
today = datetime.now()

difference_in_seconds = abs(mktime(yesterday.timetuple()) - mktime(today.timetuple()))
difference_in_minutes = difference_in_seconds / 60

Answer 15

以其他方式獲得日期之間的差異；

import dateutil.parser
import datetime
last_sent_date = "" # date string
timeDifference = current_date - dateutil.parser.parse(last_sent_date)
time_difference_in_minutes = (int(timeDifference.days) * 24 * 60) + int((timeDifference.seconds) / 60)

所以在 Min 中獲得輸出。

謝謝

Answer 16

我已經使用持續集成測試的時間差異來檢查和改進我的功能。 如果有人需要，這里是簡單的代碼

from datetime import datetime

class TimeLogger:
    time_cursor = None

    def pin_time(self):
        global time_cursor
        time_cursor = datetime.now()

    def log(self, text=None) -> float:
        global time_cursor

        if not time_cursor:
            time_cursor = datetime.now()

        now = datetime.now()
        t_delta = now - time_cursor

        seconds = t_delta.total_seconds()

        result = str(now) + ' tl -----------> %.5f' % seconds
        if text:
            result += "   " + text
        print(result)

        self.pin_time()

        return seconds


time_logger = TimeLogger()

使用：

from .tests_time_logger import time_logger
class Tests(TestCase):
    def test_workflow(self):
    time_logger.pin_time()

    ... my functions here ...

    time_logger.log()

    ... other function(s) ...

    time_logger.log(text='Tests finished')

我在日志輸出中有類似的東西

2019-12-20 17:19:23.635297 tl -----------> 0.00007
2019-12-20 17:19:28.147656 tl -----------> 4.51234   Tests finished

Answer 17

基於@Attaque 很好的答案，我提出了一個更短的日期時間差計算器的簡化版本：

seconds_mapping = {
    'y': 31536000,
    'm': 2628002.88, # this is approximate, 365 / 12; use with caution
    'w': 604800,
    'd': 86400,
    'h': 3600,
    'min': 60,
    's': 1,
    'mil': 0.001,
}

def get_duration(d1, d2, interval, with_reminder=False):
    if with_reminder:
        return divmod((d2 - d1).total_seconds(), seconds_mapping[interval])
    else:
        return (d2 - d1).total_seconds() / seconds_mapping[interval]

我對其進行了更改以避免聲明重復函數，刪除了漂亮的打印默認間隔並增加了對毫秒、周和 ISO 月份的支持（請記住，月份只是近似值，基於每個月等於365/12 ）。

其中產生：

d1 = datetime(2011, 3, 1, 1, 1, 1, 1000)
d2 = datetime(2011, 4, 1, 1, 1, 1, 2500)

print(get_duration(d1, d2, 'y', True))      # => (0.0, 2678400.0015)
print(get_duration(d1, d2, 'm', True))      # => (1.0, 50397.12149999989)
print(get_duration(d1, d2, 'w', True))      # => (4.0, 259200.00149999978)
print(get_duration(d1, d2, 'd', True))      # => (31.0, 0.0014999997802078724)
print(get_duration(d1, d2, 'h', True))      # => (744.0, 0.0014999997802078724)
print(get_duration(d1, d2, 'min', True))    # => (44640.0, 0.0014999997802078724)
print(get_duration(d1, d2, 's', True))      # => (2678400.0, 0.0014999997802078724)
print(get_duration(d1, d2, 'mil', True))    # => (2678400001.0, 0.0004999997244524721)

print(get_duration(d1, d2, 'y', False))     # => 0.08493150689687975
print(get_duration(d1, d2, 'm', False))     # => 1.019176965856293
print(get_duration(d1, d2, 'w', False))     # => 4.428571431051587
print(get_duration(d1, d2, 'd', False))     # => 31.00000001736111
print(get_duration(d1, d2, 'h', False))     # => 744.0000004166666
print(get_duration(d1, d2, 'min', False))   # => 44640.000024999994
print(get_duration(d1, d2, 's', False))     # => 2678400.0015
print(get_duration(d1, d2, 'mil', False))   # => 2678400001.4999995

Answer 18

您可能會發現這個快速片段在不太長的時間間隔內很有用：

    from datetime import datetime as dttm
    time_ago = dttm(2017, 3, 1, 1, 1, 1, 1348)
    delta = dttm.now() - time_ago
    days = delta.days # can be converted into years which complicates a bit…
    hours, minutes, seconds = map(int, delta.__format__('').split('.')[0].split(' ')[-1].split(':'))

在 Python v.3.8.6 上測試

Answer 19

這是一個易於概括或轉化為函數且合理緊湊且易於遵循的答案。

ts_start=datetime(2020, 12, 1, 3, 9, 45)
ts_end=datetime.now()
ts_diff=ts_end-ts_start
secs=ts_diff.total_seconds()
days,secs=divmod(secs,secs_per_day:=60*60*24)
hrs,secs=divmod(secs,secs_per_hr:=60*60)
mins,secs=divmod(secs,secs_per_min:=60)
secs=round(secs, 2)
answer='Duration={} days, {} hrs, {} mins and {} secs'.format(int(days),int(hrs),int(mins),secs)
print(answer)

它以Duration=270 days, 10 hrs, 32 mins and 42.13 secs形式給出答案

Answer 20

import datetime
date = datetime.date(1, 1, 1)
#combine a dummy date to the time
datetime1 = datetime.datetime.combine(date, start_time)
datetime2 = datetime.datetime.combine(date, stop_time)  
#compute the difference
time_elapsed = datetime1 - datetime2

start_time --> 日期時間對象的開始時間
end_time--> datetime 對象的結束時間

我們不能直接減去 datetime.time 對象
因此我們需要為其添加一個隨機日期（我們使用組合）
或者您可以使用“今天”而不是 (1,1,1)

希望這可以幫助